Follow up on another question I asked recently: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism

Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for every point $x \in X$ there exists an open set $U \subseteq X$ containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.

This definition is a bit alarming because it starts with "...if for every point $x \in X$ there exists an open set $U \in \mathcal{T}$...", makes it seem like a property of the underlying space. Can we always find an open $U$? But anyways.

Objective: Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism

Proof: (Honestly not sure what I am doing but proceed regardless)

Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces and function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism. We will show that $f$ is continuous and open.

First show $f$ is continuous.

$f$ is continuous if for all $V \in \mathcal{J}, f^{-1}(V) \in \mathcal{T}$. Take some $V \in \mathcal{J}$, then $V$ is a subspace equipped with subspace topology $\mathcal{J}_V = \{V \cap W| W \in \mathcal{J}\}$.

Consider the inverse of the restriction $f^{-1}|_U$ on an open set in $\mathcal{J}_V$, then $f^{-1}|_U(V \cap W) = f^{-1}|_U(V) \cap f^{-1}|_U(W) $$= f^{-1}(V) \cap U \cap f^{-1}(W) \cap U = f^{-1}(V) \cap f^{-1}(W) \cap U$.

Then $f^{-1}(V) = f^{-1}(W) \cup U \cup f^{-1}|_U(V \cap W)$. We note all the sets on the right hand side are open. In particular, $U$ is open, $f^{-1}|_U(V \cap W)$ is open by definition of homeomorphism (?? $f^{-1}(W)$ ??), hence $f$ is continuous. ($\leftarrow$ something wrong here!)

Next show $f$ is open.

$f$ is open if $\forall U \in \mathcal{T}, f(U) \in \mathcal{J}$. Consider the restriction $f|_U$ on the subspace topology on $U$, $\mathcal{T}_U = \{U \cap M | M \in \mathcal{T}\}$. $f|_U(U \cap M) = f|_U(U) \cap f|_U(M) = V \cap f(M) \cap f(U)$

Then $f$ is open since $f(U) = f|_U(U \cap M) \cup V \cup f(M)$ and $f|_U(U \cap M)$ is open by definition of homeomorphism, $V$ is open in $\mathcal{T}$ (?? $\cup f(M)$ ??) ($\Leftarrow$ another mistake here)

I'm not quite sure how to proceed with showing bijective + continuous + open + local = homeomorphism part.

Can someone help me fix those two problems and give me some ideas how to conclude that bijective local homeomorphisms are homeomorphisms?

  • $\begingroup$ "This definition is a bit alarming..." Well, it is a property of the space $X$, and of the space $Y$ and the function $f : X \to Y$. And no, you cannot always find such an open set $U$, in which case you could safely conclude that $f$ violates the definition and therefore $f$ is not a local homeomorphism. $\endgroup$
    – Lee Mosher
    Commented Jun 15, 2016 at 11:32
  • $\begingroup$ @Pom Poko In definition of local homeomorphism a condition on $f(U)$ to be open subset of $Y$ is dropped, but is used below in proofs. Is the condition really deducible from your definition? $\endgroup$ Commented Feb 25, 2018 at 20:23
  • $\begingroup$ @Pom Poko Consider an inclusion $f: *\to \{*,\{*\}\}$ of singleton in two-point anti-discrete space. $f(*)$ as a subspace of $\{*,\{*\}\}$ is isomorphic to $*$ but not open. $\endgroup$ Commented Feb 25, 2018 at 20:51

3 Answers 3


Your attempts are, unfortunately, flawed.

Since you know about local properties of $f$, it is better showing that $f$ is continuous at each point.

Let $x\in X$; we want to show that, for every open neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq V$. Let $U_x$ be an open neighborhood of $x$ and $V_x$ an open set in $Y$ such that $f$ induces a homeomorphism $f_{U_x}\colon U_x\to V_x$ and choose any open neighborhood $V$ of $f(x)$.

Then $V\cap V_x$ is an open set in $Y$ containing $f(x)$,

so there exists an open neighborhood $U$ of $x$ in $U_x$ such that $f(U)\subseteq V\cap V_x$; since $U$ is open in $U_x$ it is open in $X$ as well and $f(U)\subseteq V$ as requested.

Now you want to prove that $f$ is open. Let $A$ be open in $X$ and, for each $x\in A$, choose open sets $U_x\subseteq X$ and $V_x\subseteq Y$ so that $x\in U_x$ and $f$ induces a homeomorphism between $U_x$ and $V_x$.

For each $x\in A$, $f(U_x\cap A)$ is open in $V_x$, so it is open in $Y$ as well. Therefore $$ \bigcup_{x\in A}f(U_x\cap A) $$

equals $f(A)$ and is open in $Y$.

If $f$ is bijective, then $f^{-1}$ exists and it is continuous

because $f$ is open.

  • $\begingroup$ Excellent thank you so much. $\endgroup$
    – Fraïssé
    Commented Jun 15, 2016 at 15:57
  • $\begingroup$ If X is a singleton set then how is this true??? $\endgroup$
    – Infinity
    Commented Jan 22, 2023 at 6:13
  • $\begingroup$ @Spectrum In that case, also $Y$ is a singleton. $\endgroup$
    – egreg
    Commented Jan 22, 2023 at 9:33
  • $\begingroup$ Yes yes you are right. $\endgroup$
    – Infinity
    Commented Jan 22, 2023 at 11:50

Allow me to add another answer for proving the continuity. The difference from @egreg answer is only in looking at proof from a different angle.

Let $U \subseteq Y$ be open in $Y$. We must show that $f^{-1}(U)$ is open in $X$. Let $x \in f^{-1}(U)$ be arbitrary.

By definition of local homeomorphism, $\exists\ V_x \subseteq X$ which is a neighbourhood of $x$ such that $f(V_x)$ is open in $Y$ and $f\big\vert_{V_x}:V_x\rightarrow f(V_x)$ is a homeomorphism.

Since $U$ and $f(V_x)$ are open in $Y$, then, so is their intersection $U \cap f(V_x)$ is open in $Y$.

Also, continuity of $f\big\vert_{V_x}$ implies that,

$$f\big\vert_{V_x}^{-1}(U \cap f(V_x)) = \{x \in V_x: f(x) \in U \cap f(V_x)\} = V_x \cap f^{-1}(U)$$

is open in $X$. But $V_x \cap f^{-1}(U)$ is a neighbourhood of $x$ contained in $f^{-1}(U)$. Because $x$ is an arbitrary point in $f^{-1}(U)$, therefore,

$$f^{-1}(U) = \bigcup\limits_{x \in f^{-1}(U)}(V_x\cap f^{-1}(U))$$

is an arbitrary union of open subsets of $X$, hence, is open in $X$. Therefore, $f$ is continuous.


Claim 1: Every local homeomorphism is an open map.

Proof: Let $f:X\rightarrow Y$ be a local homeomorphism. Let $U\subseteq X$ be open. If $x\in U$ then there exists some open subset $V\subseteq X$ such that on $V$, $f$ is open onto an open subset of $Y$. Hence $f(U\cap V)$ is open in $Y$. Note, $f(x)\in f(U\cap V)\subseteq f(U)$. Thus we may conclude that $f(U)$ is the union of open sets and is therefore open.

Claim 2: Every bijective local homeomorphism is a homeomorphism.

Proof: Let $f:X\rightarrow Y$ be a bijective local homeomorphism. Let $V$ be open in $Y$. We must show that $f^{-1}(V)$ is open in $X$. Let $x\in f^{-1}(V)$. Let $U_{x}$ be an open neighborhood of $x$ such that $f|_{U_x}: U_x\rightarrow f(U_x)$ is a homeomorphism onto an open subset of $Y$. Hence, $f(x)\in f(U_x)\cap V$. Choose an open set $W_{f(x)} \owns f(x)$ such that $W_{f(x)}\subseteq f(U_x)\cap V$. Then, $f^{-1}(W_{f(x)})$ is open in $X$. Moreover, $f^{-1}(W_{f(x)})\subseteq f^{-1}(V)$ . Thus $f^{-1}(V)$ is open. Now note, if $f:X\rightarrow Y$ is a bijective local homeomorphism, then $f^{-1}: Y\rightarrow X$ is a bijective local homeomorphism. Hence repeating a similar argument as we did for $f$ allows us to conclude that $f$ is a homeomorphism. Alternatively, if $U$ is open in $X$ then $f(U)$ is open in $Y$. Hence, $(f^{-1})^{-1}(U)=f(U)$ is open in $Y$.

  • $\begingroup$ Just add some details to the post: Noting that each $V$ depends on $x$, we may denote it by $V_x$. By the proof of claim $1$, we have $f(U)=f(\cup_{x\in U} \{x\})=\cup_{x\in U} \{f(x)\}\subseteq \cup_{x\in U} f(U \cap V_x)\subseteq f(U)$. So, $f(U)=\cup_{x\in U} f(U \cap V_x)$, which is clearly open in $Y$. $\endgroup$
    – Sam Wong
    Commented Mar 5, 2021 at 6:17
  • $\begingroup$ If X is singleton then how f is an open map? $\endgroup$
    – Infinity
    Commented Jan 22, 2023 at 6:18
  • $\begingroup$ @Spectrum What? if $X$ is a singleton then its image is a singleton. Singletons arent typically open? $\endgroup$ Commented Oct 5, 2023 at 21:39

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